I was reading a source that suggested that if $X$ is a measure space (or perhaps just the interval with Lebesgue measure), and if $1 \leq p < \infty$, then any non-negative linear functional $T$ on $L^p(X)$ is continuous. ($f \geq 0$ a.e. implies $T(f) \geq 0$ as real numbers.) Is this true, or even obviously true?
I read here that positive linear functionals on $C^*$ algebras are continuous, but $L^p$ spaces don't have this structure in general… I also learned that it is true when $p = \infty$ and $X$ is a countable set with counting measure: Any positive linear functional $\phi$ on $\ell^\infty$ is a bounded linear operator and has $\|\phi \| = \phi((1,1,…)) $ … however, I don't think the same technique carries over. There is also this question, which is somewhat weaker and more general: Positive linear functional on an involutive Banach algebra
I also found some discussion here ( question (2) ): application of positive linear functionl
Best Answer
Let $X$ be any real or complex Banach space of functions or equivalence classes of functions such that with $f\in X$ also $|f|\in X$ and $\||f|\| = \|f\|$ and let $\phi$ be a positive linear functional on $X$. It is easily seen that $|\phi(f)|\le 2\phi(|f|)$ for all $f\in X$.
Suppose that $\phi$ is unbounded. Then there exists a sequence $(f_n)\subset X$ with $\|f_n\|=1$ and $|\phi(f_n)|\ge 2\cdot 4^n$. Put $g_n := |f_n|\ge 0$. Then $\|g_n\|=1$ and $\phi(g_n)\ge 4^n$ for $n\in\mathbb N$. Define the function $h := \sum_{n=1}^\infty2^{-n}g_n\in X$. Then $h\ge 2^{-n}g_n$ for each $n\in\mathbb N$ and so $\phi(h)\ge 2^{-n}\phi(g_n)\ge 2^n$ for all $n\in\mathbb N$, which is impossible.